In these notes on page 4 the author says that if a $2d$-dimensional phase space has $d$ conserved quantities $F_{\mu}$ that Poisson commute, then $H$ can be written as a function of the $F_{\mu}$. Why should this be the case?


I) Lemma: On a $2d$-dimensional symplectic manifold $(M,\{\cdot,\cdot\})$, there can at most be $d$ independent quantities that Poisson commute.

Indirect proof: Assume that $$\text{There exist }d+1 \text{ independent quantities } (F^1, \ldots , F^{d+1}) \text{ that Poisson commute.} \tag{1}$$ Consider a fixed point $p\in M$. (It does not matter which.) Define a $d+1$-dimensional subspace $$W~:=~{\rm span}_{\mathbb{R}} \{ \mathrm{d}F^1_p, \ldots, \mathrm{d}F^{d+1}_p \} ~\subseteq~T^{\ast}_pM. \tag{2}$$ The perpendicular complement $W^{\perp}$ wrt. the symplectic structure is then $d-1$ dimensional. From assumption (1) it follows that $$W ~\subseteq~ W^{\perp} \tag{2} $$ is an isotropic subspace. Contradiction. $\Box$

II) Returning to OP's question: If $H$ is not a function of the $F$'s, then there would be $d+1$ independent quantities that Poisson commute. Contradiction.

See also this related Phys.SE post.

  • $\begingroup$ I had a go at showing this, and I think I'm on the right track, but I couldn't finish it off. I've posted my work as an answer - would you mind taking a quick look at it? $\endgroup$ – preferred_anon Oct 29 '17 at 18:12
  • $\begingroup$ I updated the answer with a proof. $\endgroup$ – Qmechanic Oct 29 '17 at 19:17

Theorem: Let $M$ be a $2d$-dimensional phase space, and $F_{1},\ldots F_{d+1}:M \to \mathbb{R}$ be functionals that Poisson commute at some point $P$. Then the set $$\left\{\frac{dF_{1}}{d\mathbf{x}},\ldots \frac{dF_{d+1}}{d\mathbf{x}}\right\}$$ is linearly dependent at $p$, where $$\frac{df}{d\mathbf{x}}:=\left(\frac{\partial f}{\partial q_{1}}\ldots \frac{\partial f}{\partial p_{n}}\right)$$ and $(q_{1},\ldots q_{n},p_{1},\ldots ,p_{n})$ are coordinates on $M$.

Lemma: The set $$\left\{\frac{dF_{1}}{d\mathbf{x}},\ldots \frac{dF_{d}}{d\mathbf{x}}\right\}$$ if and only if the set $$\left\{\frac{dF_{1}}{d\mathbf{q}},\ldots \frac{dF_{d}}{d\mathbf{q}}\right\}$$ is linearly independent.

Proof of Lemma: One direction is trivial. For the other, suppose $\left\{\frac{dF_{i}}{d\mathbf{x}}\right\}$ is linearly independent, and $$\sum_{i=1}^{d}\alpha_{i}\frac{dF_{i}}{d\mathbf{q}} = 0$$ Taking the dot product, $$\sum_{i=1}^{d} \alpha_{i}\frac{dF_{i}}{d\mathbf{q}}\cdot \frac{dF_{j}}{d\mathbf{p}} = 0$$ $\{F_{i},F_{j}\} = 0$, so $$\frac{dF_{i}}{d\mathbf{p}}\cdot \frac{dF_{j}}{d\mathbf{q}} = \frac{dF_{i}}{d\mathbf{q}}\cdot \frac{dF_{j}}{d\mathbf{p}}$$ So $$\left(\sum_{i=1}^{d} \alpha_{i}\frac{dF_{i}}{d\mathbf{p}}\right)\cdot \frac{dF_{j}}{d\mathbf{q}} = 0$$ I'd like to deduce that the vector in brackets is $0$ (since then we can inherit the linear independence of $\frac{dF_{i}}{d\mathbf{x}}$). Of course, we have that it is orthogonal to $d$ vectors in $\mathbb{R}^{d}$, but if we don't know they're linearly independent I don't think we can deduce anything.


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